Let $Q=(Q_0,Q_1)$ be a quiver as shown below, where $Q_0=\{1,2\}$, and $Q_1=\{\alpha,\beta,\gamma\}$, $\alpha:1\to 1$, $\beta:1\to 2$, $\gamma:2\to 1$. I would like to show that two admissible ideals $\mathcal I_1=\langle \alpha^2-\beta\gamma,\gamma\beta-\gamma\alpha\beta,\alpha^4\rangle$ and $\mathcal I_2=\langle \alpha^2-\beta\gamma,\gamma\beta,\alpha^4\rangle$ of $KQ$ induce the same bound quiver algebra, i.e. $KQ/\mathcal I_1\cong KQ/\mathcal I_2$, where $K$ is a field with characteristic different from $2$. Still, $\mathcal I_1\neq \mathcal I_2$.
I have shown that $R_Q^4\subseteq \mathcal I_1,\mathcal I_2\subseteq R_Q^2$, so they are admissible, and it is clear that $\gamma\beta\notin \mathcal I_1$, so $\mathcal I_1\neq \mathcal I_2$. However, I am stuck now in determining the path algebra $KQ$, because there are loops and cycles in $Q$. I know that $KQ/\mathcal I$ is a basic and connected finite-dimensional $K$-algebra with an identity, having $R_Q/\mathcal I$ as radical and $\{\epsilon_1+\mathcal I,\epsilon_2+\mathcal I\}$ as a complete set of primitive orthogonal idempotents, for any admissible $\mathcal I$, since $Q$ is a finite connected quiver.
I am also wondering whether there is another way of showing the two bound quiver algebras are isomorphic, without figuring out $KQ$ itself.
Thank you very much for your help!